⚡ Divider Lab

Capacitor Voltage Divider Calculator — DC Series, AC Impedance Divider & Equal Charge Rule

Every series capacitor holds the same charge Q = Ceq·Vtotal — so voltage divides inversely with capacitance. Enter 2–4 capacitors and a total voltage: the dashboard derives Ceq, the shared charge and the stress on every capacitor, live on each keystroke.

Capacitor Voltage Divider Calculator

Live equal-charge dashboard — Ceq, shared Q and per-cap voltage stress for 2–4 series capacitors.

Number of capacitors in series

Capacitor C1

Capacitor C2

Total voltage V_total (V)

0 V400 V

Quick presets

Equal charge readout

Core Law

Q is identical on every series capacitor.

120µC

Equivalent capacitance C_eq = 1/Σ(1/Cᵢ): 5 µF

Q = C_eq × V_total: 120 µC on every plate

Self-check: defaults (2 × 10 µF @ 24 V) must read C_eq = 5 µF, Q = 120 µC, 12 V per cap.

Voltage stress per capacitor — V_n = Q / C_n

C1 · 10 µF

Highest stress12 V50%

C2 · 10 µF

12 V50%

Bars show each capacitor's share of V_total. The DC engine above computes the equal-charge split; for the 100 nF + 10 nF @ 230 V AC preset, see the AC impedance divider section below — reactance, not stored charge, sets that split.

⚡ State migration. Watch the stress move as you type: when C1 = 10 µF and C2 = 22 µF at 100 V, C1 carries 56.9 V while C2 rests at 25.9 V. Change C2 to 47 µF and the charge stays pinned, so C1's voltage climbs while C2's falls. The smallest capacitor always ends up with the highest voltage.

The governing law

The Equal Charge Rule: Why a Capacitor Divider Calculator Splits Voltage Inversely

Connect two or more capacitors in series and something quietly counterintuitive happens: the charge on every capacitor becomes identical. There is no junction between the plates where extra charge can accumulate — the inner plates of neighbouring capacitors form electrically isolated islands that can only shuttle charge from one side to the other. Push 569 µC into a string and every capacitor in it holds exactly 569 µC, whether it is a 10 µF unit or a 33 µF unit. That single fact — the equal charge rule — is the entire engine behind every capacitor divider calculation on this page, and it is why the per-capacitor voltages produced by any voltage across capacitors in series calculator must always sum back to the applied total.

Before voltages can be assigned, the string collapses into one equivalent capacitor. The combining formula is C_eq = 1/Σ(1/C_i) — the exact mirror of the parallel-resistor formula. For 10 µF, 22 µF and 33 µF in series, 1/C_eq = 0.100 + 0.0455 + 0.0303 = 0.1758 µF⁻¹, so C_eq ≈ 5.69 µF: always smaller than the smallest member of the string. Once C_eq is known, the shared charge falls out immediately as Q = C_eq × V_total, and each capacitor's individual voltage follows from its own value via V_n = Q / C_n.

Now watch the inverse relationship work as a state machine. When C1 = 10 µF and the total is 100 V, the stress on C1 alone moves to 56.90 V — more than half the supply. Swap C1 for a 4.7 µF unit and the stress on it jumps to roughly 88 V while its neighbours relax; swap it for a 100 µF unit and it coasts below 10 V while C2 and C3 pick up the difference. Every keystroke in the dashboard above replays this transition live: raise C2 from 22 µF to 47 µF and the charge stays pinned at Q = C_eq × V_total, so C1's voltage climbs while C2's falls. The stress does not spread evenly — it migrates toward the smallest plate area in the string.

❌ Don't think like a resistor divider. Voltage divides INVERSELY with capacitance in a series string — the smallest capacitor gets the highest voltage. This is the opposite of resistors (or inductors) in series, where the largest resistance gets the most voltage.

The contrast with resistors is worth restating in numbers, because the two intuitions actively fight each other. A resistive divider assigns V_n = V_total × R_n/ΣR — proportional to the element value, so the largest resistor drops the most voltage and equal resistors share equally. A capacitive divider inverts the mapping because the fixed quantity is charge rather than current: the smallest capacitor must swing the largest voltage to absorb the same charge. Carry the resistor habit into a capacitor string and you will size the smallest unit for the least stress — exactly backwards — which is why series capacitor voltage work always starts from Q, never from the ratio you would use for resistors.

Equal charge ruleCore Law

Q₁ = Q₂ = … = Q_n = C_eq × V_total

V_n = Q / C_n

C_eq = 1 / Σ(1/C_i)

One charge, three consequences: equivalent capacitance smaller than any member, voltage inversely proportional to capacitance, and per-capacitor voltages that sum exactly to the applied total.

Electronics lab verification of a capacitor voltage divider design

Same charge, unequal stressPhoto: Unsplash

Deriving V_n = Q/C_n: What a Voltage Across Capacitors in Series Calculator Computes

Collapse the string: 1/C_eq = 1/C₁ + 1/C₂ + … + 1/C_n. Units cancel cleanly when every capacitor is entered in the same unit — µF in, µF out.
Find the shared charge: Q = C_eq × V_total. With C_eq = 5.69 µF and V_total = 100 V, Q = 569 µC on every plate.
Assign each voltage: V_n = Q/C_n. The 10 µF unit takes 56.90 V, the 22 µF unit 25.86 V, and the 33 µF unit absorbs the remainder of the rail.
Verify the sum: V₁ + V₂ + V₃ must equal V_total exactly. If it does not, a pF/nF/µF conversion is almost always the culprit.

Worked example

Worked Example: 100 V Across a 10 µF, 22 µF and 33 µF Series String

Freeze one specific input state and check the live dashboard against hand arithmetic. Apply a 100 V rail across three electrolytics in series — 10 µF, 22 µF and 33 µF. The string combines to C_eq ≈ 5.6897 µF, the shared charge is Q = 5.6897 µF × 100 V ≈ 569 µC, and dividing that charge by each capacitance yields the three voltages in the table below. Every row carries the same 569 µC because the equal charge rule admits no exceptions: the inner plates cannot gain or lose net charge, they can only pass it along.

How a Series Capacitor Voltage Calculator Sums to 100 V

Series Capacitor Voltage Division — 3 Cap Example (Total V = 100V)
Capacitor	Value (µF)	Voltage Across (V)	Charge on Plate (µC)	% of Total V
C1	10	56.90	569	56.9%
C2	22	25.86	569	25.9%
C3	33	17.24	569	17.2%

Read the charge column first: 569 µC appears on every row, and that repetition is the entire proof of the equal charge rule in one glance. Then read the % of Total V column as a stress budget — the 10 µF unit is spending 56.9% of the rail on itself while the 33 µF unit spends just 17.2%. Finally, check that the Voltage Across column sums back to the 100 V you fed in; the accounting closes to the volt because charge is conserved, not because the formula was designed to look tidy. Any capacitor series voltage calculator that fails this closing check has a unit-conversion bug, not a rounding problem.

Sensitivity is the deeper lesson. When C2 = 22 µF, C3 sits at 17.24 V with 569 µC on its plates; replace C2 with a 47 µF unit and C3's share slides further down while C2 absorbs the difference; halve C1 to 4.7 µF and the hot spot migrates to C1 at roughly 88 V of the same 100 V rail. With three equal 10 µF units the split becomes a symmetric 33.3% each — the only case where the division looks intuitively "fair". Watch the same migrations happen live in the dashboard above: the bars are the state machine, and the smallest capacitor is always the state that saturates first.

Hands probing a series capacitor test circuit on an electronics workbench

Measure, then deratePhoto: Unsplash

Want the charge itself rather than the split? The capacitor charge calculator computes Q = C × V for single capacitors, and the series-parallel capacitor calculator reduces mixed networks to the single C_eq this page starts from.

Safety engineering

Capacitor Series Voltage Calculator Safety: Ratings, Derating and Balancing Resistors

The nominal division is only half the design. The other half is making sure every capacitor survives the stress the division assigns to it — including the worst case, when tolerances and leakage land unfavourably. A capacitor series voltage calculator gives you the nominal numbers; the safety margin comes from derating discipline and, at DC, from equalizing resistors that tame the leakage lottery.

Rating Limit

⚠️ Warning. For this reason, series strings of real capacitors MUST have equal or derated voltage ratings. Putting a 10V cap in series with a 35V cap can destroy the 10V cap even when the total applied voltage is below 35V.

Why does the smaller-rated unit die first? Because the division assigns voltage by capacitance, not by rating. In a 10 µF + 22 µF string across 30 V, the 10 µF unit takes 20 V and the 22 µF unit takes 10 V — so if the 10 µF part is rated 16 V and the 22 µF part is rated 35 V, the string fails at 30 V even though 30 V is below the 51 V sum of the ratings and below each individual rating in isolation. Design practice is to derate hard: run each capacitor at 50–80% of its rated voltage in the nominal calculation, then add margin for ±20% tolerance spreads and for the leakage-driven DC unbalance described below. Electrolytic aluminium capacitors deserve the most conservative margins because their leakage rises steeply near rated voltage and with age and temperature; ceramic Class-II parts drift downward in capacitance under DC bias, which silently re-weights the division after assembly.

The 10× Leakage Rule for Equalizing Resistors

At DC the steady-state distribution is set by leakage, not by charge. Once the charging transient decays, each capacitor discharges through its own leakage path, and the string settles into whatever resistive divider the leakage resistances happen to form — with the leakiest capacitor holding the least voltage and its neighbours absorbing the difference unpredictably. Equalizing (balancing) resistors wired in parallel with each capacitor override that lottery: choose each resistor so the current through it is about ten times the capacitor's worst-case leakage at working voltage. A film capacitor leaking 5 nA at 100 V needs only a 2 MΩ equalizer to dominate its leakage twenty-fold; an electrolytic leaking 50 µA at the same voltage would need roughly 200 kΩ — and that bleed path then dissipates I²R heat continuously, which is exactly why high-voltage electrolytic strings run warm idle currents and why low-leakage film strings can skip balancing entirely.

💡 Tip. Sizing equalizers: pick R so that V_rated/R ≥ 10 × I_leak(max), then verify the resistor's own power rating — a 100 kΩ resistor across a capacitor sitting at 100 V dissipates 0.1 W, so a 0.25 W part is fine but a 0.125 W part runs hot. If the standby power budget cannot afford 10×, 5× still tames most of the unbalance at half the dissipation.

Balancing resistors fix DC distribution but do nothing for the AC or transient case, where unequal capacitance values still unequalise the dV/dt stress. That is why snubber and multiplier strings combine all three mitigations: matched capacitor values, equalizing resistors, and generous derating. For the combining step — turning a messy series-parallel network into the single C_eq this page's math starts from — use the series-parallel capacitor calculator, then return here to distribute the stress.

Frequency domain

AC Capacitive Divider: Building a Voltage Divider with Capacitor Calculator Logic

Swap the DC source for an AC source and the governing variable changes from stored charge to impedance. Each capacitor now presents a reactance X_C = 1/(2πfC) measured in ohms, the same AC current flows through the series string, and the voltage divides in proportion to those reactances: V_n = V_total × X_n/ΣX. Because reactance is inversely proportional to capacitance, the LARGER capacitor drops the LESS voltage — the same direction as the equal charge rule, but reached through a different mechanism. A voltage divider with capacitor calculator built on impedance logic is frequency-aware in a way the charge rule is not: at 50 Hz a 1 µF unit looks like 3 183 Ω, at 60 Hz the identical part looks like 2 653 Ω, and every reactance in the string rescales together with 1/f.

Worked AC Example: 1 µF Over 100 nF Across 230 V at 50 Hz

Take a 1 µF capacitor in series with a 100 nF capacitor across a 230 V, 50 Hz mains-derived supply. The reactances are X_C(1 µF) = 1/(2π × 50 × 1 × 10⁻⁶) = 3 183 Ω and X_C(100 nF) = 31 831 Ω, for a series total of 35 014 Ω. The string current is 230 V ÷ 35 014 Ω = 6.57 mA, so the 1 µF unit drops 6.57 mA × 3 183 Ω ≈ 21 V and the 100 nF unit drops 6.57 mA × 31 831 Ω ≈ 209 V. The two drops sum to the full 230 V, the smaller capacitor takes ten times the stress of the larger one — exactly the C₂/C₁ ratio — and the dissipation in the string is near zero because reactance stores and returns energy rather than burning it. This is the operating principle of every capacitive dropper, and it is why mains-frequency designs must be computed at the exact working frequency: the division ratio C₂/C₁ is frequency-independent for ideal parts, but the current, the impedance level and the behaviour under real loads all move with f.

❌ Don't call it a regulator. A capacitor divider does NOT regulate — output sags with load. The moment you draw current from the tap, the load resistance parallels the lower capacitor, the effective impedance ratio shifts, and the output falls below the unloaded calculation. Capacitive dividers set a ratio for open-circuit conditions; they never hold a voltage against varying loads. For any load-dependent application you need an actual regulator or a transformer, not a reactance ratio.

Where does the AC division actually get used? In generic capacitive sensing — touch and proximity sensors read the phase shift a finger adds to a known divider — in triac snubber networks that share dV/dt stress, and in the dropper stage of transformerless supplies where a series capacitor acts as a lossless-ish current limiter. Every one of those designs starts by evaluating reactance at the working frequency, and the capacitor reactance calculator gives you X_C for any C and f pair in one step. Feed those reactances back into the proportionality above and you have a complete AC capacitive divider design flow — no simulation required for the first pass.

ℹ️ Reality check. Real capacitors also carry ESR and ESL, so above a few hundred kilohertz the reactance model under-predicts the impedance of large-value parts and the division drifts. Below resonance the error is small; near the part's self-resonant frequency the "capacitor" stops being a capacitor at all. Keep AC dividers well inside the frequency range where X_C dominates the parasitics.

Applied engineering

Practical Applications: Where a Capacitors in Series Voltage Calculator Earns Its Keep

Series capacitor strings show up anywhere a rail exceeds what a single package can stand. Triac snubber strings are the classic case: RC snubbers wired in series across a switch must share the blocking voltage, and the capacitive portion distributes that stress by the same inverse rule computed above — which is why snubber capacitors are deliberately specified with identical values and generous ratings rather than whatever is in the drawer. LED cap-drop supplies work the same territory from the AC side: a series film capacitor acts as the dropper impedance, and the "division" between that capacitor and the rest of the circuit determines the LED current — a concept application where the reactance math above, not the DC charge rule, does the real work, and where the 100 nF + 10 nF @ 230 V AC preset gives a feel for the impedance magnitudes involved.

Precision applications lean on the same math at signal levels. Crystal DC-bias networks place capacitors in series paths where the AC coupling ratio sets how much of a bias step reaches the crystal terminals; a 10:1 or 100:1 oscilloscope probe is fundamentally a divider whose capacitive arms are trimmed so the division stays flat across frequency; and high-voltage multiplier and DC-link strings distribute hundreds of volts across film capacitors whose individual ratings top out far below the bus. In every one of these, the engineering question is identical: given the string, who carries how much? A capacitors in series voltage calculator answers it in three lines of arithmetic — C_eq, Q, then V_n = Q/C_n — and the answer is the starting point for every rating, tolerance and balancing decision that follows.

Capacitor Voltage Divider Calculator Workflow: From Snubber Strings to Probe Dividers

This capacitor voltage divider calculator sits in the middle of a toolchain. Start at the Capacitor Calculator home page to decode capacitor part numbers and convert between pF, nF and µF before you enter values; browse the calculators hub for the full toolbox; reduce a mixed network with the series-parallel capacitor calculator; evaluate AC behaviour with the capacitor reactance calculator; and quantify stored charge per plate with the capacitor charge calculator. The division this page computes is the checkpoint between "the string exists" and "the string survives" — always verify it before commit.

💡 Tip. For probe-style dividers, match the time constants (R×C) of the two arms, not just the capacitance ratio — that is what keeps a 100:1 probe flat from DC to the scope's bandwidth. The capacitive arms then divide high-frequency components by C₂/C₁ while the resistive arms divide DC by R₂/R₁, and equal time constants make both divisions agree at every frequency in between.

Capacitor Voltage Divider FAQ: Eight Exact Answers

How does a capacitor voltage divider calculator work?▼

It applies the equal charge rule to whatever string you enter. The calculator first collapses your capacitors into one equivalent capacitance using Ceq = 1/Σ(1/Ci), then multiplies Ceq by the total applied voltage to get the charge Q that every plate shares, and finally divides that charge by each individual capacitance to get Vn = Q/Cn. Because the same Q appears in every row, the per-capacitor voltages always add back up to the total — for the default 2 × 10 µF at 24 V you get Ceq = 5 µF, Q = 120 µC and exactly 12 V across each capacitor. Every slider and keystroke replays the same three-step derivation live.

Why does the smallest capacitor get the most voltage?▼

Because every series capacitor is forced to hold the same charge Q, and V = Q/C makes voltage inversely proportional to capacitance at fixed charge. A 10 nF unit needs ten times more voltage than a 100 nF unit to hold the same charge, so in a series string the 10 nF part sits at ten times the electrical stress. With 100 V across 10 µF, 22 µF and 33 µF in series, the 10 µF capacitor takes 56.9 V — more than half the supply — while the 33 µF unit takes the smallest share. This inversion is the single most common cause of blown capacitors in home-built series strings.

What is the equal charge rule?▼

The equal charge rule states that capacitors connected in series carry identical charge: Q₁ = Q₂ = … = Qn = Ceq × Vtotal. The same current flows through the string during charging, and the inner plates of neighbouring capacitors form isolated islands that can only shuttle charge from one side to the other — never store extra. Each capacitor's individual voltage then follows from its own value via Vn = Q/Cn. The rule is the capacitive mirror of the equal-current rule for series resistors, and it is the reason a capacitor divider calculator can compute every output from one intermediate number.

Can I use capacitors instead of resistors for a divider?▼

Only for AC, and only for outputs that stay essentially unloaded. A capacitive divider dissipates almost no real power — an ideal capacitor loses nothing to heat — which is exactly why cap-drop LED supplies and triac snubber networks use them where a resistor divider would burn watts. But the output impedance is very high, the division ratio only holds at the design frequency, and any load resistance in parallel with the lower capacitor drags the output down. For DC the concept fails outright: after the charging transient, no current flows and the output simply sits at the rail. Use resistive dividers for measurement and bias duty, capacitive dividers for AC droppers and sensing.

How do you calculate voltage across capacitors in series?▼

Three steps. First, compute the equivalent capacitance: 1/Ceq = 1/C1 + 1/C2 + … For 10 µF, 22 µF and 33 µF that gives Ceq ≈ 5.69 µF. Second, multiply by the total voltage to get the shared charge: Q = 5.69 µF × 100 V ≈ 569 µC. Third, divide that charge by each capacitance: V1 = 569 µC ÷ 10 µF = 56.9 V, V2 = 569 µC ÷ 22 µF = 25.9 V, and V3 takes the remainder of the 100 V rail. The per-capacitor voltages must always sum to the applied total — a free self-check that your unit conversions, or your calculator, are honest.

Do I need balancing resistors across series capacitors?▼

If the string sees high DC voltage or contains electrolytics, yes. Real capacitors leak at different rates, and after the brief charging transient the leakage network — not the equal charge rule — sets the steady DC distribution: the leakiest capacitor ends up with the least voltage and its neighbours unpredictably absorb the difference. Equalizing resistors, typically sized so the bleeding current is about ten times the worst-case leakage, force a predictable resistive division on top of the capacitive one. The cost is standby dissipation and heat, which is why low-leakage film and ceramic strings at signal levels can usually skip them.

What is an AC capacitive divider?▼

An AC capacitive divider splits an alternating voltage in proportion to capacitive reactance instead of stored charge. Each capacitor presents X꜀ = 1/(2πfC) ohms, the same AC current flows through the series string, and voltage divides in proportion to those reactances — so the larger capacitor, with the smaller reactance, drops the less voltage. A 1 µF unit over 100 nF across 230 V at 50 Hz divides into roughly 21 V and 209 V. Because reactance depends on frequency, the design must be evaluated at the exact working frequency, and the output must remain essentially unloaded or the ratio collapses.

Does capacitance tolerance unbalance the division?▼

Yes, directly and proportionally. The division ratio depends only on the capacitance ratios, so ±20% tolerance parts can push roughly 8–20% extra voltage onto the smallest unit compared with the nominal calculation — and in longer strings the spread widens further. The practical consequence is that your voltage-rating margin must absorb the worst case, not the average: a string computed at exactly 100% of each rating on nominal values will exceed those ratings the moment real tolerances land unfavourably. Use 5% or better film capacitors for precision dividers, measure and match parts for high-voltage strings, and derate generously.

Keep calculating